Thermal Properties of Matter

JEE Main level, single correct, medium

81 of 126

Question 81 of 126

An experiment takes 10min⁡10\min to raise the temperature of water in a container from 0oC{0}^{o}C to 100oC{100}^{o}C and another 55min⁡55\min to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of the water to be 1cal(g∘C)−11cal{\left({g}^{^{\circ}}C\right)}^{-1} , the heat of vaporization according to this experiment will come out to be:

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Answer: (B) 550calg−1550cal{g}^{-1}

Let the constant state of supply of heat per unit time =λ=\lambda. So according to the first condition. λt1=mc×100⇒λm=100Ct1...(i)\lambda {t}_{1}=mc\times 100\Rightarrow \dfrac{\lambda }{m}=\dfrac{100C}{{t}_{1}}...(i), and according to the second condition, λt2=mLf⇒λm=Lft2...(ii)\lambda {t}_{2}=m{L}_{f}\Rightarrow \dfrac{\lambda }{m}=\dfrac{{L}_{f}}{{t}_{2}}...(ii). By (i)(i) and (ii)(ii) 100Ct1=Lft2\dfrac{100C}{{t}_{1}}=\dfrac{{L}_{f}}{{t}_{2}}⇒1cal×10010=Lf55⇒Lf=550calg−1\Rightarrow \dfrac{1cal\times 100}{10}=\dfrac{{L}_{f}}{55}\Rightarrow {L}_{f}=550cal{g}^{-1}

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